Quantum Electrodynamics

Y. D. Chong PH4401: Quantum Mechanics III

Chapter 5: Quantum Electrodynamics

This chapter gives an introduction to quantum electrodynamics, the quantum theory of the electromagnetic field and its interactions with electrons and other charged particles. We begin by formulating a quantum Hamiltonian for an electron in a classical electromagnetic field. Then we study how to quantize Maxwell’s equations, arriving at a quantum field theory in which the elementary excitations are photons—particles of light. The final step is to formulate a theory in which electrons and photons are treated on the same quantum mechanical footing, as excitations of underlying quantum fields. Along the way, we will see how relativity can be accomodated with quantum theory. Quantum electrodynamics is an extremely rich and intricate theory, and we will leave out many important topics. Interested readers are referred to are Dyson’s 1951 lecture notes on quantum electrodynamics [1], and Zee’s textbook Quantum Field Theory in a Nutshell [2].

I. QUANTIZATION OF THE LORENTZ FORCE LAW

A. Non-relativistic electrons in an electromagnetic ﬁeld

Consider a non-relativistic charged particle in an electromagnetic field. As we are mainly interested in the physics of electrons interacting with electromagnetic fields, we henceforth take the electric charge of the particle to be −e, where e = 1.602 × 10−19 C is the elementary charge. To describe particles with an arbitrary electric charge q, simply perform the substitution e → −q in the formulas you will subsequently encounter. We wish to formulate the Hamiltonian governing the quantum dynamics of such a particle, subject to two simplifying assumptions: (i) the particle has charge and mass but is otherwise “featureless” (i.e., we ignore the spin angular momentum and magnetic dipole moment that real electrons possess), and (ii) the electromagnetic field is treated as a classical field, meaning that the electric and magnetic fields are definite quantities rather than operators. (We will see how to go beyond these simplifications later.) Classically, the electromagnetic field acts on the particle via the Lorentz force law, F(r, t) = −e E(r, t) + r˙ × B(r, t) , (5.1) where r and r˙ denote the position and velocity of the particle, t is the time, and E and B are the electric and magnetic fields. If no other forces are present, Newton’s second law yields the equation of motion m¨r = −e E(r, t) + r˙ × B(r, t) , (5.2) where m is the particle’s mass. To quantize this, we must first convert the equation of motion into the form of Hamilton’s equations of motion. Let us introduce the electromagnetic scalar and vector potentials Φ(r, t) and A(r, t): ∂A E(r, t) = −∇Φ(r, t) − , (5.3) ∂t B(r, t) = ∇ × A(r, t). (5.4)

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We now postulate that the equation of motion (5.2) can be described by the Lagrangian

1 h i L(r, r˙, t) = mr˙ 2 + e Φ(r, t) − r˙ · A(r, t) . (5.5) 2

This follows the usual prescription for the Lagrangian as kinetic energy minus potential energy, with −eΦ serving as the potential energy function, except for the −er˙ · A term. To see if this Lagrangian works, plug it into the Euler-Lagrange equations ∂L d ∂L = . (5.6) ∂ri dt ∂r˙i The partial derivatives of the Lagrangian are: ∂L h i = e ∂ Φ − r˙ ∂ A ∂r i j i j i (5.7) ∂L = mr˙i − eAi. ∂r˙i

Now we want to take the total time derivative of ∂L/∂r˙i. In doing so, note that the A field has its own t-dependence, as well as varying with the particle’s t-dependent position. Thus, d ∂L d = mrï − e Ai(r(t), t) dt ∂r˙i dt (5.8) = mrï − e ∂tAi − e r˙j∂jAi.

(In the above equations, ∂i ≡ ∂/∂ri, where ri is the i-th component of the position vector, while ∂t ≡ ∂/∂t.) Plugging these expressions into the Euler-Lagrange equations (5.6) gives h i mr¨i = −e − ∂iΦ − ∂tAi +r ˙j ∂iAj − ∂jAi h i (5.9) ˙ = −e Ei(r, t) + r × B(r, t) i .

(The last step can be derived by expressing the cross product using the Levi-Cevita symbol, and using the identity εijkεlmk = δilδjm − δimδjl.) This exactly matches Eq. (5.2), as desired. We now use the Lagrangian to derive the Hamiltonian. The canonical momentum is ∂L pi = = mr˙i − eAi. (5.10) ∂r˙i The Hamiltonian is deﬁned as H(r, p) = p · r˙ − L. Using Eq. (5.10), we express it in terms of p rather than r˙:

p + eA |p + eA|2 e H = p · − + eΦ − (p + eA) · A m 2m m (5.11) |p + eA|2 e |p + eA|2 e = − A · (p + eA) − + eΦ − (p + eA) · A . m m 2m m

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After cancelling various terms, we obtain |p + eA(r, t)|2 H = − eΦ(r, t). (5.12) 2m This looks a lot like the Hamiltonian for a non-relativistic particle in a scalar potential, |p|2 H = + V (r, t). 2m In Eq. (5.12), the −eΦ term acts like a potential energy, which is no surprise. More inter- estingly, the vector potential appears via the substitution p → p + eA(r, t). (5.13)

What does this mean? Think about what “momentum” means for a charged particle in an electromagnetic field. Noether’s theorem states that each symmetry of a system (whether classical or quantum) is associated with a conservation law. Momentum is the quantity conserved when the system is symmetric under spatial translations. One of Hamilton’s equations states that dp ∂H i = , dt ∂ri which implies that if H is r-independent, then dp/dt = 0. But when the electromagnetic potentials are r-independent, the quantity mr˙ (which we usually call momentum) is not necessarily conserved! Take the potentials Φ(r, t) = 0, A(r, t) = Ctz,ˆ (5.14) where C is some constant. These potentials are r-independent, but the vector potential is time-dependent, so the −A˙ term in Eq. (5.85) gives a non-vanishing electric field: E(r, t) = −Cz,ˆ B(r, t) = 0. (5.15) The Lorentz force law then says that d (mr˙) = eCz,ˆ (5.16) dt and thus mr˙ is not conserved. On the other hand, the quantity p = mr˙ − eA is conserved: d (mr˙ − eA) = eCzˆ − eCzˆ = 0. (5.17) dt Hence, this is the appropriate canonical momentum for a particle in an electromagnetic field. We are now ready to go from classical to quantum mechanics. Replace r with the position operator ˆr, and p with the momentum operator pˆ. The resulting quantum Hamiltonian is

|pˆ + eA(ˆr, t)|2 Hˆ (t) = − eΦ(ˆr, t). (5.18) 2m

Note: the momentum operator is pˆ = −i~∇ in the wavefunction representation, as usual.

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B. Gauge symmetry

The Hamiltonian (5.18) possesses a subtle property known as gauge symmetry. Sup- pose we modify the scalar and vector potentials via the substitutions Φ(r, t) → Φ(r, t) − Λ(˙ r, t) (5.19) A(r, t) → A(r, t) + ∇Λ(r, t), (5.20) where Λ(r, t) is an arbitrary scalar field called a gauge field. This is the gauge transformation of classical electromagnetism, which as we know leaves the electric and magnetic fields unchanged. When applied to the Hamiltonian (5.18), it generates a new Hamiltonian |pˆ + eA(ˆr, t) + e∇Λ(ˆr, t)|2 Hˆ (t) = − eΦ(ˆr, t) + eΛ(˙ ˆr, t). (5.21) Λ 2m Now suppose ψ(r, t) is a wavefunction obeying the Schrödingerequation for the original Hamiltonian Hˆ : ∂ψ |pˆ + eA(ˆr, t)|2 i = Hˆ (t)ψ(r, t) = − eΦ(ˆr, t) ψ(r, t). (5.22) ~ ∂t 2m

Then it can be shown that the wavefunction ψ exp(−ieΛ/~) automatically satisfies the ˆ Schrödingerequation for the transformed Hamiltonian HΛ: ∂ ieΛ(r, t) ˆ ieΛ(r, t) i~ ψ(r, t) exp − = HΛ(t) ψ(r, t) exp − . (5.23) ∂t ~ ~ To prove this, observe how time and space derivatives act on the new wavefunction: ∂ ieΛ ∂ψ ie ieΛ ψ exp − = − Λ˙ ψ exp ∂t ∂t ~ ~ ~ (5.24) ieΛ ie ieΛ ∇ ψ exp − = ∇ψ − ∇Λ ψ exp . ~ ~ ~ When the extra terms generated by the exp(ieΛ/~) factor are slotted into the Schrödinger equation, they cancel the gauge terms in the scalar and vector potentials. For example, ieΛ h i ieΛ − i~∇ + eA + e∇Λ ψ exp − = (−i~∇ + eA) ψ exp − (5.25) ~ ~ If we apply the (−i~∇ + eA + e∇Λ) operator a second time, it has a similar effect but with the quantity in square brackets on the right-hand side of (5.25) taking the place of ψ:

2 ieΛ h 2 i ieΛ − i~∇ + eA + e∇Λ ψ exp − = |−i~∇ + eA| ψ exp − . (5.26) ~ ~ The remainder of the proof for Eq. (5.23) can be carried out straightforwardly. The above result can be stated in a simpler form if the electromagnetic ﬁelds are static. In this case, the time-independent electromagnetic Hamiltonian is |pˆ + eA(ˆr)|2 Hˆ = − eΦ(ˆr). (5.27) 2m

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ˆ Suppose H has eigenenergies {Em} and energy eigenfunctions {ψm(r)}. Then the gauge- transformed Hamiltonian |pˆ + eA(ˆr) + e∇Λ(r)|2 Hˆ = − eΦ(ˆr) (5.28) Λ 2m has the same energy spectrum {Em}, with eigenfunctions { ψm(r) exp[−ieΛ(r)/~] }.

C. The Aharonov-Bohm eﬀect

In quantum electrodynamics, it is the electromagnetic scalar and vector potentials that appear directly in the Hamiltonian, not the electric and magnetic fields. This has profound consequences. For example, even if a charged quantum particle resides in a region with zero magnetic field, it can feel the effect of nonzero vector potentials produced by magnetic fluxes elsewhere in space, a phenomenon called the Aharonov-Bohm effect. A simple setting for observing the Aharonov-Bohm effect is shown in the figure below. A particle is trapped in a ring-shaped region (an “annulus”), of radius R and width d R. Outside the annulus, we set −eΦ → ∞ so that the wavefunction vanishes; inside the annulus, we set Φ = 0. We ignore the z-dependence of all fields and wavefunctions, so that the problem is two-dimensional. We define polar coordinates (r, φ) with the origin at the ring’s center.

Now, suppose we thread magnetic ﬂux (e.g., using a solenoid) through the origin, which lies in the region enclosed by the annulus. This ﬂux can be described via the vector potential Φ A(r, φ) = B e , (5.29) 2πr φ

where eφ is the unit vector pointing in the azimuthal direction. We can verify from Eq. (5.29) that the total magnetic flux through any loop of radius r enclosing the origin is (ΦB/2πr)(2πr) = ΦB. The fact that this is independent of r implies that the magnetic flux density is concentrated in an infintesimal area surrounding the origin, and zero everywhere else. However, the vector potential A is nonzero everywhere. The time-independent Schrödingerequation is

2 1 eΦB −i~∇ + eφ ψ(r, φ) = Eψ(r, φ), (5.30) 2m 2πr

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with the boundary conditions ψ(R ± d/2, 0) = 0. For suﬃciently large R, we can guess that the eigenfunctions have the form ( ψ cos π (r − R) eikRφ, r ∈ [R − d/2,R + d/2] ψ(r, φ) ≈ 0 d (5.31) 0 otherwise.

This describes a “waveguide mode” with a half-wavelength wave proﬁle in the r direction (so as to vanish at r = R ± d/2), traveling in the azimuthal direction with wavenumber k. The normalization constant ψ0 is unimportant. We need the wavefunction to be single-valued under a 2π variation in the azimuthal coordinate, so n k · 2πR = 2πn ⇒ k = , where n ∈ . (5.32) R Z Plugging this into Eq. (5.30) yields the energy levels " # 1 n eΦ 2 π 2 E = ~ + B + ~ (5.33) n 2m R 2πR d e2 nh2 π2 2 = Φ + + ~ . (5.34) 8π2mR2 B e 2md2

These energy levels are sketched versus the magnetic ﬂux ΦB in the ﬁgure below:

Each energy level has a quadratic dependence on ΦB. Variations in ΦB aﬀect the energy levels despite the fact that B = 0 in the annular region where the electron resides. This is a manifestation of the Aharonov-Bohm eﬀect.

It is noteworthy that the curves of different n are centered at different values of ΦB corresponding to multiples of h/e = 4.13567 × 10−5 T m2, a fundamental unit of magnetic flux called the magnetic flux quantum. In other words, changing ΦB by an exact multiple of h/e leaves the energy spectrum unchanged! This invariance property, which does not depend on the width of the annulus or any other geometrical parameters of the system, can be explained using gauge symmetry. When an extra flux of nh/e (where n ∈ Z) is

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threaded through the annulus, Eq. (5.29) tells us that the change in vector potential is ∆A = (n~/er)eφ. But we can undo the eﬀects of this via the gauge ﬁeld ( n ∇Λ = −(n /er) e Λ(r, φ) = − ~ φ ⇒ ~ φ (5.35) e e−ieΛ/~ = einφ.

Note that this Λ is not single-valued, but that’s not a problem! Both ∇Λ and the phase factor exp(−ieΛ/~) are single-valued, and those are the quantities that enter into the gauge symmetry relations (5.19)–(5.20).

II. DIRAC’S THEORY OF THE ELECTRON

A. The Dirac Hamiltonian

So far, we have been using p2/2m-type Hamiltonians, which are limited to describing non-relativistic particles. In 1928, Paul Dirac formulated a Hamiltonian that can describe electrons moving close to the speed of light, thus successfully combining quantum theory with special relativity. Another triumph of Dirac’s theory is that it accurately predicts the magnetic moment of the electron. Dirac’s theory begins from the time-dependent Schr¨odingerwave equation, ˆ i~ ∂t ψ(r, t) = Hψ(r, t). (5.36)

Note that the left side has a ﬁrst-order time derivative. On the right, the Hamiltonian Hˆ contains spatial derivatives in the form of momentum operators. We know that time and space derivatives of wavefunctions are related to energy and momentum by

i~ ∂t ↔ E, −i~ ∂j ↔ pj. (5.37) We also know that the energy and momentum of a relativistic particle are related by

3 2 2 4 X 2 2 E = m c + pj c , (5.38) j=1 where m is the rest mass and c is the speed of light. Note that E and p appear to the same order in this equation. (Following the usual practice in relativity theory, we use Roman indices j ∈ {1, 2, 3} for the spatial coordinates {x, y, z}.) Since the left side of the Schrödingerequation (5.36) has a first-order time derivative, a relativistic Hamiltonian should involve first-order spatial derivatives. So we make the guess

3 ˆ 2 X H = α0mc + αjpˆjc, (5.39) j=1

2 wherep ˆj ≡ −i~∂/∂xj. The mc and c factors are placed for later convenience. We now need to determine the dimensionless “coeﬃcients” α0, α1, α2, and α3.

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For a wavefunction with deﬁnite momentum p and energy E,

3 ! ˆ 2 X Hψ = Eψ ⇒ α0mc + αjpjc ψ = E ψ. (5.40) j=1

This is obtained by replacing thep ˆj operators with deﬁnite numbers. If ψ is a scalar, this 2 P would imply that α0mc + j αjpjc = E for certain scalar coeﬃcients {α0, . . . , α3}, which does not match the relativistic energy-mass-momentum relation (5.38). But we can get things to work if ψ(r, t) is a multi-component wavefunction, rather than a scalar wavefunction, and the α’s are matrices acting on those components via the matrix- vector product operation. In that case,

3 ˆ 2 X H =α ˆ0mc + αˆjpˆjc, wherep ˆj ≡ −i~ ∂j, (5.41) j=1

where the hats on {αˆ0,..., αˆ3} indicate that they are matrix-valued. Applying the Hamil- tonian twice gives 3 !2 2 X 2 αˆ0mc + αˆjpjc ψ = E ψ. (5.42) j=1 This can be satisﬁed if 3 !2 2 X 2 ˆ αˆ0mc + αˆjpjc = E I, (5.43) j=1 where Iˆ is the identity matrix. Expanding the square (and taking care of the fact that the αˆµ matrices need not commute) yields

2 2 4 X 3 X 2 2 ˆ αˆ0m c + (ˆα0αˆj +α ˆjαˆ0) mc pj + αˆjαˆj0 pjpj0 c = E I. (5.44) j jj0

This reduces to Eq. (5.38) if theα ˆµ matrices satisfy

αˆ2 = Iˆ for µ = 0, 1, 2, 3, and µ (5.45) αˆµαˆν +α ˆναˆµ = 0 for µ 6= ν.

(We use Greek symbols for indices ranging over the four spacetime coordinates {0, 1, 2, 3}.) The above can be written more concisely using the anticommutator:

{αˆµ, αˆν} = 2δµν, for µ, ν = 0, 1, 2, 3. (5.46) ˆ Also, we need theα ˆµ matrices to be Hermitian, so that H is Hermitian. It turns out that the smallest possible Hermitian matrices that can satisfy Eq. (5.46) are 4 × 4 matrices. The choice of matrices (or “representation”) is not uniquely determined.

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One particularly useful choice is called the Dirac representation: ˆ I 0ˆ 0ˆ σ ˆ1 αˆ0 = , αˆ1 = 0ˆ −Iˆ σˆ1 0ˆ (5.47) 0ˆ σ ˆ2 0ˆ σ ˆ3 αˆ2 = , αˆ3 = , σˆ2 0ˆ σˆ3 0ˆ where {σˆ1, σˆ2, σˆ3} denote the usual Pauli matrices. Since theα ˆµ’s are 4 × 4 matrices, it follows that ψ(r) is a four-component ﬁeld.

B. Eigenstates of the Dirac Hamiltonian

According to Eq. (5.38), the energy eigenvalues of the Dirac Hamiltonian are s 2 4 X 2 2 E = ± m c + pj c . (5.48) j

This is plotted below:

The energy spectrum forms two hyperbolic bands. For each p, there are two degenerate positive energy eigenvalues, and two degenerate negative energy eigenvalues, for a total of four eigenvalues (matching the number of wavefunction components). The upper band matches the dispersion relation for a massive relativistic particle, as desired. But what about the negative-energy band? Who ordered that? It might be possible for us to ignore the existence of the negative-energy states, if we only ever consider an isolated electron; we could just declare the positive-energy states to be the ones we are interested in, and ignore the others. However, the problem becomes hard to dismiss once we let the electron interact with another system, such as the electromagnetic ﬁeld. Under such circumstances, the availability of negative-energy states extending down to E → −∞ would destabilize the positive-energy electron states, since the electron can repeatedly hop to states with ever more negative energies by shedding energy (e.g., by emitting photons). This is obviously problematic. However, let us wait for a while (till SectionIID) to discuss how the stability problem might be resolved. For now, let us take a closer look at the meaning of the Dirac wavefunction. Its four components represent a four-fold “internal” degree of freedom, distinct from the electron’s

94 Y. D. Chong PH4401: Quantum Mechanics III ordinary kinematic degrees of freedom. Since there are two energy bands, the assignment of an electron to the upper or lower band (or some superposition thereof) consitutes two degrees of freedom. Each band must then posssess a two-fold degree of freedom (so that 2 × 2 = 4), which turns out to be associated with the electron’s spin.

To see explicitly how this works, let us pick a representation for theα ˆµ matrices. The choice of representation determines how the four degrees of freedom are encoded in the individual wavefunction components. We will use the Dirac representation (5.47). In this case, it is convenient to divide the components into upper and lower parts, ψ (r, t) ψ(r, t) = A , (5.49) ψB(r, t) where ψA and ψB have two components each. Then, for an eigenstate with energy E and momentum p, applying (5.47) to the Dirac equation (5.41) gives 1 X ψ = σˆ p ψ , (5.50) A E − mc2 j j B j 1 X ψ = σˆ p ψ . (5.51) B E + mc2 j j A j Consider the non-relativistic limit, |p| → 0, for which E approaches either mc2 or −mc2. For 2 the upper band (E & mc ), the vanishing of the denominator in Eq. (5.50) tells us that the 2 wavefunction is dominated by ψA. Conversely, for the lower band (E . −mc ), Eq. (5.51) tells us that the wavefunction is dominated by ψB. We can thus associate the upper (A) and lower (B) components with the band degree of freedom. Note, however, that this is only an approximate association that holds in the non-relativistic limit! In the relativistic regime, upper-band states can have non-vanishing values in the B components, and vice versa. (There does exist a way to make the upper/lower spinor components correspond rigorously to positive/negative energies, but this requires a more complicated representation than the Dirac representation [3].)

C. Dirac electrons in an electromagnetic ﬁeld

To continue pursuing our objective of interpreting the Dirac wavefunction, we must determine how the electron interacts with an electromagnetic ﬁeld. We introduce electromagnetism by following the same procedure as in the non-relativistic theory (SectionIA): add −eΦ(r, t) as a scalar potential function, and add the vector potential via the substitution pˆ → pˆ + eA(ˆr, t). (5.52) Applying this recipe to the Dirac Hamiltonian (5.41) yields ( ) 2 X h i i~ ∂tψ = αˆ0mc − eΦ(r, t) + αˆj − i~ ∂j + eAj(r, t) c ψ(r, t). (5.53) j You can check that this has the same gauge symmetry properties as the non-relativistic theory discussed in SectionIB.

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In the Dirac representation (5.47), Eq. (5.53) reduces to

2 X i~ ∂t ψA = + mc − eΦ ψA + σˆj − i~∂j + eAj c ψB (5.54) j 2 X i~ ∂t ψB = − mc − eΦ ψB + σˆj − i~∂j + eAj c ψA, (5.55) j where ψA and ψB are the previously-introduced two-component objects corresponding to the upper and lower halves of the Dirac wavefunction. In the non-relativistic limit, solutions to the above equations can be cast in the form mc2 ψA(r, t) = ΨA(r, t) exp −i t ~ (5.56) mc2 ψB(r, t) = ΨB(r, t) exp −i t . ~ The exponentials on the right side are the exp(−iωt) factor corresponding to the rest energy mc2, which dominates the electron’s energy in the non-relativistic limit. (Note that by using +mc2 rather than −mc2, we are explicitly referencing the positive-energy band.) If the electron is in an eigenstate with p = 0 and there are no electromagnetic ﬁelds, ΨA and ΨB would just be constants. Now suppose the electron is non-relativistic but not in a p = 0 eigenstate, and the electromagnetic ﬁelds are weak but not necessarily vanishing. In that case, ΨA and ΨB are functions that vary with t, but slowly. Plugging this ansatz into Eqs. (5.54)–(5.55) gives X i~ ∂t ΨA = −eΦΨA + σˆj − i~∂j + eAj c ΨB (5.57) j 2 X i~ ∂t + 2mc + eΦ ΨB = σˆj − i~∂j + eAj c ΨA. (5.58) j On the left side of Eq. (5.58), the 2mc2 term dominates over the other two, so 1 X Ψ ≈ σˆ − i ∂ + eA Ψ . (5.59) B 2mc j ~ j j A j Plugging this into Eq. (5.57) yields ( ) 1 X i ∂ Ψ = −eΦ + σˆ σˆ − i ∂ + eA − i ∂ + eA Ψ . (5.60) ~ t A 2m j k ~ j j ~ k k A jk ˆ P Using the identityσ ˆjσˆk = δjkI + i i εijkσi: ( 1 2 i ∂t ΨA = − eΦ + − i ∇ + eA ~ 2m ~ ) (5.61) i X + ε σˆ − i ∂ + eA − i ∂ + eA Ψ . 2m ijk i ~ j j ~ k k A ijk

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Look carefully at the last term in the curly brackets. Expanding the square yields

i X ε σˆ − ∂ ∂ − i e∂ A − i eA ∂ + A ∂ + e2A A . 2m ijk i j k ~ j k ~ k j j k j k ijk

Due to the antisymmetry of εijk, all terms inside the parentheses that are symmetric under j and k cancel out when summed over. The only survivor is the second term, which gives

e X e ~ ε σˆ ∂ A = ~ σˆ · B(r, t), (5.62) 2m ijk i j k 2m ijk where B = ∇ × A is the magnetic ﬁeld. Hence, 1 2 ~e i ∂t ΨA = −eΦ + − i ∇ + eA − − σˆ · B ΨA. (5.63) ~ 2m ~ 2m

This is an exact match for Eq. (5.18), except that the Hamiltonian has an additional term of the form −µˆ · Bˆ . This additional term corresponds to the potential energy of a magnetic dipole of moment µ in a magnetic field B. The Dirac theory therefore predicts the electron’s magnetic dipole moment to be e |µ| = ~ . (5.64) 2m Remarkably, this matches the experimentally-observed magnetic dipole moment to about one part in 103. The residual mismatch between Eq. (5.64) and the actual magnetic dipole moment of the electron is understood to arise from quantum fluctuations of the electronic and electromagnetic quantum fields. Using the full theory of quantum electrodynamics, that “anomalous magnetic moment” can also be calculated and matches experiment to around one part in 109, making it one of the most precise theoretical predictions in physics! For details, see Ref. [2]. It is noteworthy that we did not set out to include spin in the theory, yet it arose, seemingly unavoidably, as a by-product of formulating a relativistic theory of the electron. This is a manifestation of the general principle that relativistic quantum theory is more constrained than non-relativistic quantum theory [1]. Due to the demands imposed by relativistic symmetries, spin is not allowed to be an optional part of the theory of the relativistic electron—it has to be built into the theory at a fundamental level.

D. Positrons and Dirac Field Theory

As noted in SectionIIB, the stability of the quantum states described by the Dirac equation is threatened by the presence of negative-energy solutions. To get around this problem, Dirac suggested that what we regard as the “vacuum” may actually be a state, called the Dirac sea, in which all negative-energy states are occupied. Since electrons are fermions, the Pauli exclusion principle would then forbid decay into the negative-energy states, stabilizing the positive-energy states. At ﬁrst blush, the idea seems ridiculous; how can the vacuum contain an inﬁnite number of particles? However, we shall see that the idea becomes more plausible if the Dirac

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equation is reinterpreted as a single-particle construction which arises from a more fundamental quantum field theory. The Dirac sea idea is an inherently multi-particle concept, and we know from Chapter 4 that quantum field theory is a natural framework for describing multi-particle quantum states. Let us therefore develop this theory. Consider again the eigenstates of the single-particle Dirac Hamiltonian with definite mo- menta and energies. Denote the positive-energy wavefunctions by

u eik·r kσ = hr|k, +, σi, where Hˆ |k, +, σi = |k, +, σi. (5.65) (2π)3/2 kσ

The negative-energy wavefunctions are

v e−ik·r kσ = hr|k, −, σi, where Hˆ |k, −, σi = − |k, −, σi. (5.66) (2π)3/2 kσ

Note that |k, −, σi denotes a negative-energy eigenstate with momentum −~k, not ~k. The reason for this notation, which uses diﬀerent symbols to label the positive-energy and negative-energy states, will become clear later. Each of the ukσ and vkσ terms are four- component objects (spinors), and for any given k, the set

{ukσ, vk,σ | σ = 1, 2}

forms an orthonormal basis for the four-dimensional spinor space. Thus,

X n ∗ n X n ∗ n (ukσ) ukσ0 = δσσ0 , (ukσ) vkσ0 = 0, etc. (5.67) n n

n Here we use the notation where ukσ is the n-th component of the ukσ spinor, and likewise for the v’s. Following the second quantization procedure from Chapter 4, let us introduce a fermionic Fock space HF , as well as a set of creation/annihilation operators:

ˆ† ˆ bkσ and bkσ create/annihilate |k, +, σi ˆ† ˆ dkσ and dkσ create/annihilate |k, −, σi. These obey the fermionic anticommutation relations

ˆ ˆ† 3 0 ˆ ˆ† 3 0 {bkσ, b 0 0 } = δ (k − k ) δσσ0 , {dkσ, d 0 0 } = δ (k − k ) δσσ0 k σ k σ (5.68) ˆ ˆ ˆ ˆ ˆ ˆ {bkσ, bk0σ0 } = {bkσ, dk0σ0 } = {dkσ, dk0σ0 } = 0, etc.

The Hamiltonian is Z ˆ 3 X ˆ† ˆ ˆ† ˆ H = d k kσ bkσbkσ − dkσdkσ , (5.69) σ and applying the annihilation operators to the vacuum state |∅i gives zero: ˆ ˆ bkσ|∅i = dkσ|∅i = 0. (5.70)

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When formulating bosonic field theory, we defined a local field annihilation operator that annihilates a particle at a given point r. In the infinite-system limit, this took the form Z ˆ 3 ψ(r) = d k ϕk(r)a ˆk, (5.71)

ˆ ˆ† 0 3 0 and the orthonormality of the ϕk wavefunctions implied that [ψ(r), ψ (r )] = δ (r − r ). Similarly, we can use the Dirac Hamiltonian’s eigenfunctions (5.65)–(5.66) to deﬁne

Z d3k X ψˆ (r) = un eik·r ˆb + vn e−ik·r dˆ . (5.72) n (2π)3/2 kσ kσ kσ kσ σ Note that there are two terms in the parentheses because the positive-energy and negative- energy states are denoted by differently-labeled annihilation operators. Moreover, since the wavefunctions are four-component spinors, the field operators have a spinor index n. Using the spinor orthonormality conditions (5.67) and the anticommutation relations (5.68), we can show that n ˆ ˆ† 0 o 3 0 ψn(r), ψn0 (r ) = δnn0 δ (r − r ), (5.73) ˆ with all other anticommutators vanishing. Hence, ψn(r) can be regarded as an operator that annihilates a four-component fermion at point r. Now let us define the operators ˆ† cˆkσ = dkσ. (5.74) Using these, the fermionic anticommutation relations can be re-written as

ˆ ˆ† 3 0 † 3 0 {bkσ, b 0 0 } = δ (k − k ) δσσ0 , {cˆkσ, cˆ 0 0 } = δ (k − k ) δσσ0 k σ k σ (5.75) ˆ ˆ ˆ {bkσ, bk0σ0 } = {bkσ, cˆk0σ0 } = {cˆkσ, cˆk0σ0 } = 0, etc.

† Hencec ˆkσ andc ˆkσ formally satisfy the criteria to be regarded as creation and annihilation † operators. The particle created byc ˆkσ is called a positron, and is equivalent to the absence of a d-type particle (i.e., a negative-energy electron). The Hamiltonian (5.69) can now be written as Z ˆ 3 X ˆ† ˆ † H = d k kσ bkσbkσ +c ˆkσcˆkσ + constant, (5.76) σ which explicitly shows that the positrons have positive energies (i.e., the absence of a negative-energy particle is equivalent to the presence of a positive-energy particle). With † further analysis, which we will skip, it can be shown that the positron created byc ˆkσ has positive charge e and momentum ~k. The latter is thanks to the deﬁnition adopted in Eq. (5.66); the absence of a momentum −~k particle is equivalent to the presence of a momentum ~k particle. As for the ﬁeld annihilation operator (5.72), it can be written as

Z d3k X ψˆ (r) = un eik·r ˆb + vn e−ik·r cˆ† . (5.77) n (2π)3/2 kσ kσ kσ kσ σ

99 Y. D. Chong PH4401: Quantum Mechanics III

The c-type annihilation operators do not annihilate |∅i. However, let us deﬁne

0 Y ˆ† |∅ i = dkσ|∅i, (5.78) kσ which is evidently a formal description of the Dirac sea state. Then

0 ˆ† Y ˆ† cˆkσ|∅ i = dkσ dk0σ0 |∅i = 0. (5.79) k0σ0

At the end of the day, we can regard the quantum field theory as being defined in terms of b-type and c-type operators, using the anticommutators (5.75), the Hamiltonian (5.76), and the field operator (5.77), along with the vacuum state |∅0i. The elementary particles in this theory are electrons and positrons with strictly positive energies. The single-particle Dirac theory, with its quirky negative-energy states, can then be interpreted as a special construct that maps the quantum field theory into single-particle language. Even though we actually started from the single-particle description, it is the quantum field theory, and its vacuum state |∅0i, that is more fundamental. There are many more details about the Dirac theory that we will not discuss here. One particularly important issue is how the particles transform under Lorentz boosts and other changes in coordinate system. For such details, the reader is referred to Ref. [1].

III. QUANTIZING THE ELECTROMAGNETIC FIELD

Previously (Chapter 4, Sec. IV.C), we have gone through the process of quantizing a scalar boson field. The classical field is decomposed into normal modes, and each mode is quantized by assigning it an independent set of creation and annihilation operators. By comparing the oscillator energies in the classical and quantum regimes, we can derive the Hermitian operator corresponding to the classical field variable, expressed using the creation and annihilation operators. We will use the same approach, with only minor adjustments, to quantize the electromagnetic field. First, consider a “source-free” electromagnetic field—i.e., with no electric charges and currents. Without sources, Maxwell’s equations (in SI units, and in a vacuum) reduce to:

∇ · E = 0 (5.80) ∇ · B = 0 (5.81) ∂B ∇ × E = − (5.82) ∂t 1 ∂E ∇ × B = . (5.83) c2 ∂t Once again, we introduce the scalar potential Φ and vector potential A: ∂A E = −∇Φ − (5.84) ∂t B = ∇ × A. (5.85)

100 Y. D. Chong PH4401: Quantum Mechanics III

With these relations, Eqs. (5.81) and (5.82) are satisﬁed automatically via vector identities. The two remaining equations, (5.80) and (5.83), become:

∂ ∇2Φ = − ∇ · A (5.86) ∂t 1 ∂2 1 ∂ ∇2 − A = ∇ Φ + ∇ · A . (5.87) c2 ∂t2 c2 ∂t

In the next step, we choose a convenient gauge called the Coulomb gauge:

Φ = 0, ∇ · A = 0. (5.88)

(To see that we can always make such a gauge choice, suppose we start out with a scalar potential Φ0 and vector potential A0 not satisfying (5.88). Perform a gauge transformation R t 0 0 ˙ with a gauge ﬁeld Λ(r, t) = − dt Φ0(r, t ). The new scalar potential is Φ = Φ0 + Λ = 0; moreover, the new vector potential satisﬁes

Z t 2 0 2 0 ∇ · A = ∇ · A0 − ∇ Λ = ∇ · A0 + dt ∇ Φ0(r, t ). (5.89)

Upon using Eq. (5.86), we find that ∇ · A = 0.) In the Coulomb gauge, Eq. (5.86) is automatically satisfied. The sole remaining equation, (5.87), simplifies to 1 ∂2 ∇2 − A = 0. (5.90) c2 ∂t2 This has plane-wave solutions of the form A(r, t) = A ei(k·r−ωt) + c.c. e, (5.91) where A is a complex number (the mode amplitude) that specifies the magnitude and phase of the plane wave, e is a real unit vector (the polarization vector) that specifies which direction the vector potential points along, and “c.c.” denotes the complex conjugate of the first term. Referring to Eq. (5.90), the angular frequency ω must satisfy

ω = c|k|. (5.92)

Moreover, since ∇ · A = 0, it must be the case that

k · e = 0. (5.93)

In other words, the polarization vector is perpendicular to the propagation direction. For any given k, we can choose (arbitrarily) two orthogonal polarization vectors. Now suppose we put the electromagnetic ﬁeld in a box of volume V = L3, with periodic boundary conditions (we will take L → ∞ at the end). The k vectors form a discrete set: 2πn k = j , n ∈ Z, for j = 1, 2, 3. (5.94) j L j

101 Y. D. Chong PH4401: Quantum Mechanics III

Then the vector potential field can be decomposed as a superposition of plane waves, X i(k·r−ωkt) A(r, t) = Akλ e + c.c. ekλ, where ωk = c|k|. (5.95) kλ Here, λ is a two-fold polarization degree of freedom indexing the two possible orthogonal polarization vectors for each k. (We won’t need to specify how exactly these polarization vectors are defined, so long as the definition is used consistently.) To convert the classical field theory into a quantum field theory, for each (k, λ) we define an independent set of creation and annihilation operators:

† † † aˆkλ, aˆk0λ0 = δkk0 δλλ0 , aˆkλ, aˆk0λ0 = aˆkλ, aˆk0λ0 = 0. (5.96) Then the Hamiltonian for the electromagnetic ﬁeld is

ˆ X † H = ~ωk aˆkλaˆkλ, where ωk = c|k|. (5.97) kλ The vector potential is now promoted into a Hermitian operator in the Heisenberg picture: ˆ X i(k·r−ωkt) A(r, t) = Ckλ aˆkλ e + h.c. ekλ. (5.98) kλ

Here, Ckλ is a constant to be determined, and “h.c.” denotes the Hermitian conjugate. The creation and annihilation operators in this equation are Schr¨odinger picture (t = 0) operators. The particles they create/annihilate are photons—elementary particles of light.

To ﬁnd Ckλ, we compare the quantum and classical energies. Suppose the electromagnetic ﬁeld is in a coherent state |αi such that for any k and λ,

aˆkλ|αi = αkλ|αi (5.99)

for some αkλ ∈ C. From this and Eq. (5.98), we identify the corresponding classical ﬁeld X i(k·r−ωkt) A(r, t) = Akλ e + c.c. ekλ, where Ckλαkλ = Akλ. (5.100) kλ For each k and λ, Eqs. (5.84)–(5.85) give the electric and magnetic ﬁelds i(k·r−ωkt) Ekλ = iωkAkλ e + c.c. ekλ (5.101) i(k·r−ωkt) Bkλ = iAkλ e + c.c. k × ekλ. (5.102)

In the classical theory of electromagnetism, Poynting’s theorem tells us that the total energy carried by a classical plane electromagnetic wave is Z 3 0 2 2 2 E = d r Ekλ + c Bkλ V 2 (5.103) 2 2 = 2 0 ωk |Akλ| V.

102 Y. D. Chong PH4401: Quantum Mechanics III

Here, V is the volume of the enclosing box, and we have used the fact that terms like e2ik·r vanish when integrated over r. Hence, we make the correspondence r 2 2 2 ~ 2 0 ωk |Ckλαkλ| V = ~ωk|αkλ| ⇒ Ckλ = . (5.104) 20ωkV

We thus arrive at the result

ˆ X † H = ~ωk aˆkλaˆkλ kλ r ˆ X ~ i(k·r−ωkt) (5.105) A(r, t) = aˆkλ e + h.c. ekλ 20ωkV kλ † ωk = c|k|, aˆkλ, aˆk0λ0 = δkk0 δλλ0 , aˆkλ, aˆk0λ0 = 0.

To describe infinite free space rather than a finite-volume box, we take the L → ∞ limit and re-normalize the creation and annihilation operators by the replacement r (2π)3 aˆ → aˆ . (5.106) kλ V kλ Then the sums over k become integrals over the infinite three-dimensional space:

Z ˆ 3 X † H = d k ~ωk aˆkλaˆkλ λ r Z (5.107) ˆ 3 X ~ i(k·r−ωkt) A(r, t) = d k 3 aˆkλ e + h.c. ekλ 16π 0ωk λ † 3 0 ωk = c|k|, aˆkλ, aˆk0λ0 = δ (k − k )δλλ0 , aˆkλ, aˆk0λ0 = 0.

IV. THE ELECTRON-PHOTON INTERACTION

Having derived quantum theories for the electron and the electromagnetic ﬁeld, we can put them together to describe how electrons interact with the electromagnetic ﬁeld by absorbing and/or emitting photons. Here, we present the simplest such calculation.

Let He be the Hilbert space for one electron, and HEM be the Hilbert space for the electromagnetic ﬁeld. The combined system is thus described by He ⊗ HEM. We seek a Hamiltonian of the form H = He + HEM + Hint, (5.108)

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where He is the Hamiltonian for the “bare” electron, HEM is the Hamiltonian for the source- free electromagnetic field, and Hint is an interaction Hamiltonian describing how the electron interacts with photons. Let us once again adopt the Coulomb gauge, so that the scalar potential is zero, and the electromagnetic field is solely described via the vector potential. In SectionIA, we saw that the effect of the vector potential on a charged particle can be described via the substitution

pˆ → pˆ + eA(ˆr, t). (5.109)

In SectionIIC, we saw that this substitution is applicable not just to non-relativistic particles, but also to fully relativistic particles described by the Dirac Hamiltonian. Previously, we have treated the A in this substitution as a classical object lacking quantum dynamics of its own. Now, we replace it by the vector potential operator derived in SectionIII:

 r X ~ i(k·r−ωkt)  aˆkλ e + h.c. ekλ, (finite volume)  20ωkV Aˆ (ˆr, t) = kλ (5.110) Z r  3 X ~ i(k·ˆr−ωkt)  d k 3 aˆkλ e + h.c. ekλ, (infinite space).  16π 0ωk λ Using this, together with either the electronic and electromagnetic Hamiltonians, we can finally describe the photon emission process. Suppose a non-relativistic electron is orbiting an atomic nucleus in an excited state |1i ∈ He. Initially, the photon field is in its vacuum state |∅i ∈ HEM. Hence, the initial state of the combined system is

|ψii = |1i ⊗ |∅i. (5.111)

Let Hint be the Hamiltonian term responsible for photon absorption/emission. If Hint = 0, then |ψii would be an energy eigenstate. The atom would remain in its excited state forever.

In actuality, Hint is not zero, so |ψii is not an energy eigenstate. As the system evolves, the excited electron may decay into its ground state |0i by emitting a photon with energy E, equal to the energy diﬀerence between the atom’s excited state |1i and ground state |0i. For a non-relativistic electron, the Hamiltonian (5.18) yields the interaction Hamiltonian e H = pˆ · Aˆ + h.c. , (5.112) int 2m

where Aˆ must now be treated as a ﬁeld operator, not a classical ﬁeld.

Consider the states that |ψii can decay into. There is a continuum of possible ﬁnal states, each having the form (kλ) † |ψf i = |0i ⊗ aˆkλ|∅i , (5.113) which describes the electron being in its ground state and the electromagnetic ﬁeld containing one photon, with wave-vector k and polarization λ. According to Fermi’s Golden Rule (see Chapter 2), the decay rate is

2π 2 (kλ) ˆ κ = hψf |Hint|ψii D(E), (5.114) ~

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where (··· ) denotes the average over the possible decay states of energy E (i.e., equal to the energy of the initial state), and D(E) is the density of states. (kλ) ˆ To calculate the matrix element hψf |Hint|ψii, let us use the infinite-volume version of the vector field operator (5.110). (You can check that using the finite-volume version yields the same results; see Exercise 2.) We will use the Schrödingerpicture operator, equivalent to setting t = 0 in Eq. (5.110). Then Z r (kλ) ˆ e 3 0 X ~ hψf |Hint|ψii = d k 3 2m 16π 0ωk0 jλ (5.115) −ik·ˆr −ik·ˆr j † × h0|pˆje |1i + 0|e pˆj|1 ek0λ0 h∅|aˆkλaˆk0λ0 |∅i.

† 3 0 We can now use the fact that h∅|aˆkλaˆk0λ0 |∅i = δ (k − k )δλλ0 . Moreover, we approximate the exp(−ik·ˆr) factors in the brakets with 1; this is a good approximation since the size of a −9 −6 typical atomic orbital (. 10 m) is much smaller than the optical wavelength (∼ 10 m), meaning that exp(−ik · r) does not vary appreciably over the range of positions r where the orbital wavefunctions are signiﬁcant. The above equation then simpliﬁes to r (kλ) e X hψ |Hˆ |ψ i ≈ ~ h0|pˆ |1i ej . (5.116) f int i m 16π3 ω j kλ j 0 k

ˆ 2 We can make a further simpliﬁcation by observing that for He = |pˆ| /2m + V (r),

ˆ imEd [He, ˆr] = −i~p/m ⇒ h0|pˆj|1i = − . (5.117) ~ The complex number d = h0|r|1i, called the transition dipole moment, is easily calculated from the orbital wavefunctions. Thus, r (kλ) ˆ E hψf |Hint|ψii ≈ −ie 3 d · ekλ. (5.118) 16π 0 2 (kλ) ˆ 2 e E 2 hψf |Hint|ψii ≈ 3 d · ekλ . (5.119) 16π 0 (Check for yourself that Eq. (5.119) should, and does, have units of [E2V ].) We now need the average over the possible photon states (k, λ). In taking this average, the polarization vector runs over all possible directions, and a standard angular integration shows that

3 3 X X 1 |d|2 |d · e |2 = |d |2 e2 = |d |2 · = . (5.120) kλ j j j 3 3 j=1 j=1

The last thing we need is the density of photon states. Using the dispersion relation E = ~c|k|, we can show (see Exercise 3) that 8πE2 D(E) = . (5.121) ~3c3

105 Y. D. Chong PH4401: Quantum Mechanics III

This includes a factor of 2 for the photons’ two-fold polarization degree of freedom. Putting everything together, we arrive at the following decay rate:

e2 E3 |d|2 κ = 4 3 (5.122) 3π~ c 0 We can make this look nicer by deﬁning the dimensionless ﬁne-structure constant

e2 1 α = ≈ , (5.123) 4π0~c 137 and deﬁning ω = E/~ as the frequency of the emitted photon. The resulting decay rate is

4αω3 |d|2 κ = . (5.124) 3c2

The ﬁgure below compares this prediction to experimentally-determined decay rates for the simplest excited states of hydrogen, lithium, and sodium atoms. The experimental data are derived from atomic emission line-widths, and correspond to the rate of spontaneous emission (also called the “Einstein A coeﬃcient”) as the excited state decays to the ground state. For the Fermi’s Golden Rule curve, we simply approximated the transition dipole moment as |d| ≈ 10−10 m (based on the fact that |d| has units of length, and the length scale of an atomic orbital is about an angstrom); to be more precise, d ought to be calculated using the actual orbital wavefunctions. Even with the crude approximations we have made, the predictions are within striking distance of the experimental values.

FIG. 1: Spontaneous emission rates (Einstein A coeﬃcients) for the 2p → 1s transition in hydrogen, the 2p → 2s transition in lithium, and the 3p → 3s transition in sodium. Data points extracted from the NIST Atomic Spectra Database (https://www.nist.gov/pml/atomic-spectra-database). The dashed curve shows the decay rate based on Fermi’s Golden Rule, with |d| ≈ 10−10 m.

106 Y. D. Chong PH4401: Quantum Mechanics III

Exercises

1. In SectionIII, we derived the vector potential operator, in an inﬁnite volume, to be

Z r ˆ 3 X ~ i(k·r−ωkt) A(r, t) = d k 3 aˆkλ e + h.c. ekλ. (5.125) 16π 0ωk λ

† 3 0 Since [âkλ, aˆk0λ0 ] = δ (k − k )δλλ0 , the creation and annihilation operators each have units of [x3/2]. Prove that Aˆ has the same units as the classical vector potential. 2. Repeat the spontaneous decay rate calculation from SectionIV using the finite- volume versions of the creation/annihilation operators and the vector potential operator (5.110). Show that it yields the same result (5.123). 3. The density of photon states at energy E is defined as Z 3 D(E) = 2 d k δ(E − Ek), (5.126)

where Ek = ~c|k|. Note the factor of 2 accounting for the polarizations. Prove that 8πE2 D(E) = , (5.127) ~3c3 and show that D(E) has units of [E−1V −1].